100 results found for "quadratic-equation" in Class 10.
यदि किसी द्विघात समीकरण की जड़ें \(2+\sqrt{5}\) और \(2-\sqrt{5}\) हैं, तो समीकरण कौन-सा है?
If the roots of a quadratic equation are \(2+\sqrt{5}\) and \(2-\sqrt{5}\), which is the equation?
#quadratic-roots
#forming-equation
#surd-roots
A \(x^2-4x-1=0\)
B \(x^2+4x-1=0\)
C \(x^2-4x+1=0\)
D \(x^2+4x+1=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-4x-1=0\)
Step 1
Concept
\(The sum of roots is (4) and the product is (-1). Use (x^2-(\)sum)x+product\(=0) to get the answer.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-4x-1=0). The sum of roots is (4) and the product is (-1). Use (x^2-(\)sum)x+product\(=0) to get the answer.\)
Step 3
Exam Tip
जड़ों का योग (4) और गुणनफल (-1) है। \(समीकरण (x^2-(\)योग)x+गुणनफल\(=0) से उत्तर मिलता है\)।
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यदि किसी द्विघात समीकरण की जड़ें एक-दूसरे की व्युत्क्रम हैं और उनका योग \(\frac{5}{2}\) है, तो समीकरण कौन-सा हो सकता है?
If the roots of a quadratic equation are reciprocals of each other and their sum is \(\frac{5}{2}\), which equation is possible?
#quadratic-roots
#reciprocal-roots
#forming-equation
A \(2x^2-5x+2=0\)
B \(2x^2+5x+2=0\)
C \(x^2-5x+2=0\)
D \(5x^2-2x+2=0\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2-5x+2=0\)
Step 1
Concept
Reciprocal roots have product (1). Multiplying \(x^2-\frac{5}{2}x+1=0\) by (2) gives \(2x^2-5x+2=0\).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2-5x+2=0\). Reciprocal roots have product (1). Multiplying \(x^2-\frac{5}{2}x+1=0\) by (2) gives \(2x^2-5x+2=0\).
Step 3
Exam Tip
व्युत्क्रम जड़ों का गुणनफल (1) होता है। समीकरण \(x^2-\frac{5}{2}x+1=0\) को (2) से गुणा करने पर \(2x^2-5x+2=0\) मिलता है।
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यदि किसी द्विघात समीकरण की जड़ों का योग (7) और गुणनफल (10) है, तो समीकरण कौन-सा है?
If the sum of roots of a quadratic equation is (7) and the product is (10), which is the equation?
#quadratic-roots
#forming-equation
#sum-product
A \(x^2+7x+10=0\)
B \(x^2-10x+7=0\)
C \(x^2-7x+10=0\)
D \(x^2+10x-7=0\)
Explanation opens after your attempt
Correct Answer
C. \(x^2-7x+10=0\)
Step 1
Concept
\(The equation is (x^2-(\)sum)x+product\(=0). Hence (x^2-7x+10=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is C. (x^2-7x+10=0). The equation is (x^2-(\)sum)x+product\(=0). Hence (x^2-7x+10=0) is correct.\)
Step 3
Exam Tip
\(जड़ों का समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2-7x+10=0) सही है\)।
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यदि किसी द्विघात समीकरण की जड़ें (3) और (-2) हैं और \(x^2\) का गुणांक (2) है, तो समीकरण कौन-सा है?
If the roots of a quadratic equation are (3) and (-2), and the coefficient of \(x^2\) is (2), which is the equation?
#quadratic-roots
#forming-equation
#leading-coefficient
A \(2x^2-2x-12=0\)
B \(2x^2+2x-12=0\)
C \(2x^2-2x+12=0\)
D \(x^2-x-6=0\)
Explanation opens after your attempt
Correct Answer
A. \(2x^2-2x-12=0\)
Step 1
Concept
The monic equation is \(x^2-x-6=0\). Since the coefficient of \(x^2\) must be (2), multiply the whole equation by (2).
Step 2
Why this answer is correct
The correct answer is A. \(2x^2-2x-12=0\). The monic equation is \(x^2-x-6=0\). Since the coefficient of \(x^2\) must be (2), multiply the whole equation by (2).
Step 3
Exam Tip
मॉनिक समीकरण \(x^2-x-6=0\) है। \(x^2\) का गुणांक (2) चाहिए, इसलिए पूरे समीकरण को (2) से गुणा करें।
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यदि किसी मोनिक द्विघात समीकरण के दोनों मूल (5) और (5) हैं तो समीकरण कौन सा है?
If both roots of a monic quadratic equation are (5) and (5), which equation is it?
#roots
#equal_roots
#equation_from_roots
A \(x^2-10x+25=0\)
B \(x^2+10x+25=0\)
C \(x^2-5x+25=0\)
D \(x^2+5x-25=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-10x+25=0\)
Step 1
Concept
If both roots are (5), the equation is ((x-5)2 =0). Expanding it gives \(x^2-10x+25=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-10x+25=0\). If both roots are (5), the equation is ((x-5)2 =0). Expanding it gives \(x^2-10x+25=0\).
Step 3
Exam Tip
दोनों मूल (5) हों तो समीकरण ((x-5)2 =0) होगा। इसे खोलने पर \(x^2-10x+25=0\) मिलता है।
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यदि किसी मोनिक द्विघात समीकरण के दोनों मूल (-4) और (-4) हैं तो समीकरण कौन सा है?
If both roots of a monic quadratic equation are (-4) and (-4), which equation is it?
#roots
#equal_roots
#equation_from_roots
A \(x^2+8x+16=0\)
B \(x^2-8x+16=0\)
C \(x^2+4x+16=0\)
D \(x^2-4x-16=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+8x+16=0\)
Step 1
Concept
If both roots are (-4), the equation is ((x+4)2 =0). Expanding it gives \(x^2+8x+16=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+8x+16=0\). If both roots are (-4), the equation is ((x+4)2 =0). Expanding it gives \(x^2+8x+16=0\).
Step 3
Exam Tip
दोनों मूल (-4) हों तो समीकरण ((x+4)2 =0) होगा। इसे खोलने पर \(x^2+8x+16=0\) मिलता है।
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यदि किसी मोनिक द्विघात समीकरण के दोनों मूल (7) और (7) हैं तो समीकरण कौन सा है?
If both roots of a monic quadratic equation are (7) and (7) then which equation is it?
#roots
#equal_roots
#equation_from_roots
A \(x^2-14x+49=0\)
B \(x^2+14x+49=0\)
C \(x^2-7x+49=0\)
D \(x^2+7x-49=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-14x+49=0\)
Step 1
Concept
With both roots (7) we get ((x-7)2 =0) which is \(x^2-14x+49=0\). Form a perfect square from repeated roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-14x+49=0\). With both roots (7) we get ((x-7)2 =0) which is \(x^2-14x+49=0\). Form a perfect square from repeated roots.
Step 3
Exam Tip
दोनों मूल (7) होने पर ((x-7)2 =0) मिलता है जो \(x^2-14x+49=0\) है। दोहराए मूल से पूर्ण वर्ग बनाएं।
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यदि (4) और (9) किसी द्विघात समीकरण के मूल हैं, तो वह समीकरण कौनसा हो सकता है?
If (4) and (9) are roots of a quadratic equation, which equation can it be?
#quadratic
#construct-equation
#roots
A \(x^2-13x+36=0\)
B \(x^2+13x+36=0\)
C \(x^2-36x+13=0\)
D \(x^2+36x+13=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-13x+36=0\)
Step 1
Concept
If roots are (4) and (9), then ((x-4)(x-9)=0), that is \(x^2-13x+36=0\). In exams, form factors with opposite signs of roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-13x+36=0\). If roots are (4) and (9), then ((x-4)(x-9)=0), that is \(x^2-13x+36=0\). In exams, form factors with opposite signs of roots.
Step 3
Exam Tip
मूल (4) और (9) हों तो ((x-4)(x-9)=0), यानी \(x^2-13x+36=0\) है। परीक्षा में मूलों के विपरीत चिन्ह से गुणनखंड बनाएं।
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यदि (3) और (7) किसी द्विघात समीकरण के मूल हैं, तो वह समीकरण कौनसा हो सकता है?
If (3) and (7) are roots of a quadratic equation, which equation can it be?
#quadratic
#construct-equation
#roots
A \(x^2-10x+21=0\)
B \(x^2+10x+21=0\)
C \(x^2-21x+10=0\)
D \(x^2+21x+10=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-10x+21=0\)
Step 1
Concept
If roots are (3) and (7), then ((x-3)(x-7)=0), that is \(x^2-10x+21=0\). In exams, form factors with opposite signs of roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-10x+21=0\). If roots are (3) and (7), then ((x-3)(x-7)=0), that is \(x^2-10x+21=0\). In exams, form factors with opposite signs of roots.
Step 3
Exam Tip
मूल (3) और (7) हों तो ((x-3)(x-7)=0), यानी \(x^2-10x+21=0\) है। परीक्षा में मूलों के विपरीत चिन्ह से गुणनखंड बनाएं।
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यदि (2) और (5) किसी द्विघात समीकरण के मूल हैं, तो वह समीकरण कौनसा हो सकता है?
If (2) and (5) are roots of a quadratic equation, which equation can it be?
#quadratic
#construct-equation
#roots
A \(x^2-7x+10=0\)
B \(x^2+7x+10=0\)
C \(x^2-10x+7=0\)
D \(x^2+10x+7=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-7x+10=0\)
Step 1
Concept
If the roots are (2) and (5), the equation is ((x-2)(x-5)=0), that is \(x^2-7x+10=0\). In exams, form factors with opposite signs of roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-7x+10=0\). If the roots are (2) and (5), the equation is ((x-2)(x-5)=0), that is \(x^2-7x+10=0\). In exams, form factors with opposite signs of roots.
Step 3
Exam Tip
मूल (2) और (5) हों तो समीकरण ((x-2)(x-5)=0) यानी \(x^2-7x+10=0\) है। परीक्षा में मूलों के विपरीत चिन्ह से गुणनखंड बनाएं।
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\(\frac{1}{2}\) और \(\frac{3}{4}\) जड़ों वाला द्विघात समीकरण कौन-सा है?
Which quadratic equation has roots \(\frac{1}{2}\) and \(\frac{3}{4}\)?
#quadratic-roots
#forming-equation
#fractional-roots
A \(8x^2-10x+3=0\)
B \(8x^2+10x+3=0\)
C \(4x^2-10x+3=0\)
D \(8x^2-6x+3=0\)
Explanation opens after your attempt
Correct Answer
A. \(8x^2-10x+3=0\)
Step 1
Concept
The sum is \(\frac{5}{4}\) and the product is \(\frac{3}{8}\). Multiply \(x^2-\frac{5}{4}x+\frac{3}{8}=0\) by (8).
Step 2
Why this answer is correct
The correct answer is A. \(8x^2-10x+3=0\). The sum is \(\frac{5}{4}\) and the product is \(\frac{3}{8}\). Multiply \(x^2-\frac{5}{4}x+\frac{3}{8}=0\) by (8).
Step 3
Exam Tip
जड़ों का योग \(\frac{5}{4}\) और गुणनफल \(\frac{3}{8}\) है। समीकरण \(x^2-\frac{5}{4}x+\frac{3}{8}=0\) को (8) से गुणा करें।
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यदि किसी द्विघात समीकरण में (a=2), (b=-7), (c=3) हैं, तो समीकरण कौन-सा है?
If a quadratic equation has (a=2), (b=-7), (c=3), which equation is it?
#quadratic-equations
#forming-equation
#coefficients
#medium
A \(2x^2+7x+3=0\)
B \(2x^2-7x+3=0\)
C \(3x^2-7x+2=0\)
D \(2x^2-3x+7=0\)
Explanation opens after your attempt
Correct Answer
B. \(2x^2-7x+3=0\)
Step 1
Concept
Substituting in \(ax^2+bx+c=0\) gives \(2x^2-7x+3=0\). Keep the negative sign of (b).
Step 2
Why this answer is correct
The correct answer is B. \(2x^2-7x+3=0\). Substituting in \(ax^2+bx+c=0\) gives \(2x^2-7x+3=0\). Keep the negative sign of (b).
Step 3
Exam Tip
मानक रूप \(ax^2+bx+c=0\) में मान रखने पर \(2x^2-7x+3=0\) मिलता है। (b) का ऋण चिन्ह साथ रखें।
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कौन सा समीकरण शुद्ध द्विघात समीकरण है?
Which equation is a pure quadratic equation?
#quadratic equations
#pure quadratic
#recognition
A \(4x^2-16=0\)
B \(x^2+5x+6=0\)
C (x+16=0)
D \(x^3-16=0\)
Explanation opens after your attempt
Correct Answer
A. \(4x^2-16=0\)
Step 1
Concept
A pure quadratic has no linear (x) term. \(4x^2-16=0\) is such an equation.
Step 2
Why this answer is correct
The correct answer is A. \(4x^2-16=0\). A pure quadratic has no linear (x) term. \(4x^2-16=0\) is such an equation.
Step 3
Exam Tip
शुद्ध द्विघात में (x) वाला रैखिक पद नहीं होता। \(4x^2-16=0\) ऐसा समीकरण है।
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सामान्य द्विघात समीकरण \(ax^2+bx+c=0\) में जड़ें समान हों, तो सही संबंध कौन-सा है?
For the general quadratic equation \(ax^2+bx+c=0\), which relation is correct when the roots are equal?
#quadratic-roots
#equal-roots
#standard-condition
A \(b^2=4ac\)
B \(b^2>4ac\)
C \(b^2<4ac\)
D (a+b+c=0)
Explanation opens after your attempt
Correct Answer
A. \(b^2=4ac\)
Step 1
Concept
For equal real roots, the discriminant \(D=b^2-4ac=0\). Therefore \(b^2=4ac\) is the correct relation.
Step 2
Why this answer is correct
The correct answer is A. \(b^2=4ac\). For equal real roots, the discriminant \(D=b^2-4ac=0\). Therefore \(b^2=4ac\) is the correct relation.
Step 3
Exam Tip
समान वास्तविक जड़ों के लिए विविक्तकर \(D=b^2-4ac=0\) होता है। इसलिए \(b^2=4ac\) सही संबंध है।
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यदि \(x^2-7x+12=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(2\alpha+3\) और \(2\beta+3\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-7x+12=0\), which equation has roots \(2\alpha+3\) and \(2\beta+3\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-20x+99=0\)
B \(x^2-14x+99=0\)
C \(x^2-20x+91=0\)
D \(x^2+20x+99=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-20x+99=0\)
Step 1
Concept
The original roots are (3) and (4). The new roots are (9) and (11), so the equation is \(x^2-20x+99=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-20x+99=0\). The original roots are (3) and (4). The new roots are (9) and (11), so the equation is \(x^2-20x+99=0\).
Step 3
Exam Tip
मूल जड़ें (3) और (4) हैं। नई जड़ें (9) और (11) हैं, इसलिए समीकरण \(x^2-20x+99=0\) है।
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यदि \(x^2-5x+6=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha+\beta\) और \(\alpha\beta\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are roots of \(x^2-5x+6=0\), which equation has roots \(\alpha+\beta\) and \(\alpha\beta\)?
#quadratic-roots
#new-equation
#sum-product-roots
A \(x^2-11x+30=0\)
B \(x^2+11x+30=0\)
C \(x^2-5x+6=0\)
D \(x^2-30x+11=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-11x+30=0\)
Step 1
Concept
Here \(\alpha+\beta=5\) and \(\alpha\beta=6\). The new roots are (5) and (6), so the equation is \(x^2-11x+30=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-11x+30=0\). Here \(\alpha+\beta=5\) and \(\alpha\beta=6\). The new roots are (5) and (6), so the equation is \(x^2-11x+30=0\).
Step 3
Exam Tip
\(\alpha+\beta=5\) और \(\alpha\beta=6\) हैं। नई जड़ें (5) और (6) हैं, इसलिए समीकरण \(x^2-11x+30=0\) है।
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यदि \(x^2-9x+14=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha-3\) और \(\beta-3\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-9x+14=0\), which equation has roots \(\alpha-3\) and \(\beta-3\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-3x-4=0\)
B \(x^2+3x-4=0\)
C \(x^2-3x+4=0\)
D \(x^2-9x+14=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-3x-4=0\)
Step 1
Concept
The original roots are (2) and (7). The new roots are (-1) and (4), so the equation is \(x^2-3x-4=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-3x-4=0\). The original roots are (2) and (7). The new roots are (-1) and (4), so the equation is \(x^2-3x-4=0\).
Step 3
Exam Tip
मूल जड़ें (2) और (7) हैं। नई जड़ें (-1) और (4) होंगी, इसलिए समीकरण \(x^2-3x-4=0\) है।
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यदि \(x^2-6x+5=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(3\alpha-2\) और \(3\beta-2\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-6x+5=0\), which equation has roots \(3\alpha-2\) and \(3\beta-2\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-14x+13=0\)
B \(x^2-18x+45=0\)
C \(x^2-14x+25=0\)
D \(x^2+14x+13=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-14x+13=0\)
Step 1
Concept
The original roots are (1) and (5), so the new roots are (1) and (13). Their equation is \(x^2-14x+13=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-14x+13=0\). The original roots are (1) and (5), so the new roots are (1) and (13). Their equation is \(x^2-14x+13=0\).
Step 3
Exam Tip
मूल जड़ें (1) और (5) हैं, इसलिए नई जड़ें (1) और (13) हैं। उनका समीकरण \(x^2-14x+13=0\) है।
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यदि \(x^2-3x-2=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\alpha^2,\beta^2\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(x^2-3x-2=0\), which equation has roots \(\alpha^2,\beta^2\)?
#quadratic-roots
#squared-roots
#new-equation
A \(x^2-13x+4=0\)
B \(x^2+13x+4=0\)
C \(x^2-9x+4=0\)
D \(x^2-13x-4=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-13x+4=0\)
Step 1
Concept
Here \(\alpha+\beta=3\) and \(\alpha\beta=-2\). Thus \(\alpha^2+\beta^2=13\) and \(\alpha^2\beta^2=4\), so the equation is \(x^2-13x+4=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-13x+4=0\). Here \(\alpha+\beta=3\) and \(\alpha\beta=-2\). Thus \(\alpha^2+\beta^2=13\) and \(\alpha^2\beta^2=4\), so the equation is \(x^2-13x+4=0\).
Step 3
Exam Tip
\(\alpha+\beta=3\) और \(\alpha\beta=-2\) है। इसलिए \(\alpha^2+\beta^2=13\) और \(\alpha^2\beta^2=4\), अतः समीकरण \(x^2-13x+4=0\) है।
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\(x^2-5x+6=0\) की जड़ें \(\alpha,\beta\) हैं। \(\alpha+1,\beta+1\) जड़ों वाला समीकरण कौन-सा है?
The roots of \(x^2-5x+6=0\) are \(\alpha,\beta\). Which equation has roots \(\alpha+1,\beta+1\)?
#quadratic-roots
#transformed-roots
#new-equation
A \(x^2-7x+12=0\)
B \(x^2-5x+12=0\)
C \(x^2-6x+8=0\)
D \(x^2+7x+12=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-7x+12=0\)
Step 1
Concept
The original roots are (2) and (3), so the new roots are (3) and (4). Their equation is \(x^2-7x+12=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-7x+12=0\). The original roots are (2) and (3), so the new roots are (3) and (4). Their equation is \(x^2-7x+12=0\).
Step 3
Exam Tip
मूल जड़ें (2) और (3) हैं, इसलिए नई जड़ें (3) और (4) होंगी। उनका समीकरण \(x^2-7x+12=0\) है।
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यदि \(3x^2-10x+3=0\) की जड़ें \(\alpha,\beta\) हैं, तो \(\frac{1}{\alpha},\frac{1}{\beta}\) जड़ों वाला समीकरण कौन-सा है?
If \(\alpha,\beta\) are the roots of \(3x^2-10x+3=0\), which equation has roots \(\frac{1}{\alpha},\frac{1}{\beta}\)?
#quadratic-roots
#reciprocal-roots
#new-equation
A \(3x^2-10x+3=0\)
B \(3x^2+10x+3=0\)
C \(x^2-10x+3=0\)
D \(10x^2-3x+3=0\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2-10x+3=0\)
Step 1
Concept
Here \(\alpha+\beta=\frac{10}{3}\) and \(\alpha\beta=1\). The reciprocal roots also have sum \(\frac{10}{3}\) and product (1).
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-10x+3=0\). Here \(\alpha+\beta=\frac{10}{3}\) and \(\alpha\beta=1\). The reciprocal roots also have sum \(\frac{10}{3}\) and product (1).
Step 3
Exam Tip
यहाँ \(\alpha+\beta=\frac{10}{3}\) और \(\alpha\beta=1\) है। व्युत्क्रम जड़ों का योग \(\frac{10}{3}\) और गुणनफल (1) ही रहता है।
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यदि किसी द्विघात समीकरण के मूल (c) और (-c) हैं तो अचर पद और अग्र गुणांक के अनुपात \(\frac{d}{a}\) का मान क्या होगा?
If the roots of a quadratic equation are (c) and (-c), what is the value of the ratio \(\frac{d}{a}\) of constant term to leading coefficient?
#roots
#opposite_roots
#product
A \(-c^2\)
B \(c^2\)
C (0)
D (2c)
Explanation opens after your attempt
Correct Answer
A. \(-c^2\)
Step 1
Concept
\(\frac{d}{a}\) is the product of roots. Here (c(-c)=-c-2 ).
Step 2
Why this answer is correct
The correct answer is A. \(-c^2\). \(\frac{d}{a}\) is the product of roots. Here (c(-c)=-c-2 ).
Step 3
Exam Tip
\(\frac{d}{a}\) मूलों का गुणनफल होता है। यहां (c(-c)=-c-2 ) है।
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यदि द्विघात समीकरण के मूल (6) और \(\frac{1}{6}\) हैं तो उनके बारे में सही कथन कौन सा है?
If the roots of a quadratic equation are (6) and \(\frac{1}{6}\), which statement is correct about them?
#roots
#reciprocal_roots
#concept
A वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other
B वे बराबर मूल हैं / They are equal roots
C उनका योग (1) है / Their sum is (1)
D उनका गुणनफल (0) है / Their product is (0)
Explanation opens after your attempt
Correct Answer
A. वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other
Step 1
Concept
\(6\cdot\frac{1}{6}=1\), so the roots are reciprocals. Reciprocal roots have product (1).
Step 2
Why this answer is correct
The correct answer is A. वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other. \(6\cdot\frac{1}{6}=1\), so the roots are reciprocals. Reciprocal roots have product (1).
Step 3
Exam Tip
\(6\cdot\frac{1}{6}=1\) है इसलिए दोनों व्युत्क्रम मूल हैं। व्युत्क्रम मूलों का गुणनफल (1) होता है।
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यदि (x=3) और (x=8) किसी मोनिक द्विघात समीकरण के मूल हैं तो (x) का गुणांक क्या होगा?
If (x=3) and (x=8) are roots of a monic quadratic equation, what will be the coefficient of (x)?
#roots
#coefficient_from_roots
#monic
A (-11)
B (11)
C (24)
D (-24)
Explanation opens after your attempt
Step 1
Concept
\(The sum of roots is (3+8=11). In a monic equation the coefficient of (x) is (-(\)sum\()=-11).\)
Step 2
Why this answer is correct
\(The correct answer is A. (-11). The sum of roots is (3+8=11). In a monic equation the coefficient of (x) is (-(\)sum\()=-11).\)
Step 3
Exam Tip
मूलों का योग (3+8=11) है। मोनिक समीकरण में (x) का गुणांक (-(योग\()=-11) होता है\)।
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यदि किसी द्विघात समीकरण के मूलों का योग (0) है तो मूलों के बारे में सही कथन कौन सा हो सकता है?
If the sum of roots of a quadratic equation is (0), which statement about the roots can be correct?
#roots
#zero_sum
#reasoning
A मूल एक दूसरे के विपरीत हैं / The roots are opposites of each other
B दोनों मूल हमेशा (1) हैं / Both roots are always (1)
C दोनों मूल हमेशा धनात्मक हैं / Both roots are always positive
D मूलों का गुणनफल हमेशा (0) है / The product is always (0)
Explanation opens after your attempt
Correct Answer
A. मूल एक दूसरे के विपरीत हैं / The roots are opposites of each other
Step 1
Concept
If \(\alpha+\beta=0\), then \(\beta=-\alpha\). Therefore the roots can be opposites.
Step 2
Why this answer is correct
The correct answer is A. मूल एक दूसरे के विपरीत हैं / The roots are opposites of each other. If \(\alpha+\beta=0\), then \(\beta=-\alpha\). Therefore the roots can be opposites.
Step 3
Exam Tip
यदि \(\alpha+\beta=0\) है तो \(\beta=-\alpha\) होता है। इसलिए मूल विपरीत हो सकते हैं।
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यदि किसी द्विघात समीकरण में (a=1), (b=-20), (c=100) है तो मूल कौन से होंगे?
If a quadratic equation has (a=1), (b=-20), and (c=100), what will be the roots?
#roots
#coefficients
#repeated_root
A (10) और (10) / (10) and (10)
B (-10) और (-10) / (-10) and (-10)
C (20) और (100) / (20) and (100)
D (0) और (10) / (0) and (10)
Explanation opens after your attempt
Correct Answer
A. (10) और (10) / (10) and (10)
Step 1
Concept
The equation is \(x^2-20x+100=0\), which becomes ((x-10)2 =0). Therefore both roots are (10).
Step 2
Why this answer is correct
The correct answer is A. (10) और (10) / (10) and (10). The equation is \(x^2-20x+100=0\), which becomes ((x-10)2 =0). Therefore both roots are (10).
Step 3
Exam Tip
समीकरण \(x^2-20x+100=0\) है जो ((x-10)2 =0) बनता है। इसलिए दोनों मूल (10) हैं।
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यदि (x+6) और (x-2) किसी द्विघात समीकरण के गुणनखंड हैं तो उसके मूल कौन से हैं?
If (x+6) and (x-2) are factors of a quadratic equation, what are its roots?
#roots
#factors_to_roots
#zero_product
A (-6) और (2) / (-6) and (2)
B (6) और (-2) / (6) and (-2)
C (6) और (2) / (6) and (2)
D (-6) और (-2) / (-6) and (-2)
Explanation opens after your attempt
Correct Answer
A. (-6) और (2) / (-6) and (2)
Step 1
Concept
From (x+6=0), (x=-6), and from (x-2=0), (x=2). The sign changes when finding a root from a factor.
Step 2
Why this answer is correct
The correct answer is A. (-6) और (2) / (-6) and (2). From (x+6=0), (x=-6), and from (x-2=0), (x=2). The sign changes when finding a root from a factor.
Step 3
Exam Tip
(x+6=0) से (x=-6) और (x-2=0) से (x=2) मिलता है। गुणनखंड से मूल निकालते समय चिन्ह बदलता है।
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यदि द्विघात समीकरण के मूल (4) और (-9) हैं तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) का मान क्या होगा?
If the roots of a quadratic equation are (4) and (-9), what is the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\)?
#roots
#reciprocal_sum
#identity
A \(\frac{5}{36}\)
B -\(\frac{5}{36}\)
C \(\frac{13}{36}\)
D -\(\frac{13}{36}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{5}{36}\)
Step 1
Concept
Here \(\alpha+\beta=-5\) and \(\alpha\beta=-36\). Therefore \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-36}=\frac{5}{36}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5}{36}\). Here \(\alpha+\beta=-5\) and \(\alpha\beta=-36\). Therefore \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-36}=\frac{5}{36}\).
Step 3
Exam Tip
यहां \(\alpha+\beta=-5\) और \(\alpha\beta=-36\) है। इसलिए \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-36}=\frac{5}{36}\) है।
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यदि मूलों का योग (11) और गुणनफल (28) है तो मोनिक द्विघात समीकरण कौन सा होगा?
If the sum of roots is (11) and product is (28), which monic quadratic equation is formed?
#roots
#equation_from_sum_product
#monic
A \(x^2-11x+28=0\)
B \(x^2+11x+28=0\)
C \(x^2-28x+11=0\)
D \(x^2+28x+11=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-11x+28=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-11x+28=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-11x+28=0). A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-11x+28=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2-11x+28=0) सही है\)।
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यदि किसी द्विघात समीकरण के लिए (D=81) है तो वास्तविक मूलों की प्रकृति क्या होगी?
If (D=81) for a quadratic equation, what will be the nature of real roots?
#roots
#discriminant
#nature
A दो भिन्न वास्तविक मूल / Two distinct real roots
B दो बराबर वास्तविक मूल / Two equal real roots
C कोई वास्तविक मूल नहीं / No real root
D एक मूल (0) / One root (0)
Explanation opens after your attempt
Correct Answer
A. दो भिन्न वास्तविक मूल / Two distinct real roots
Step 1
Concept
Since (D=81>0), there will be two distinct real roots. (D>0) indicates different roots.
Step 2
Why this answer is correct
The correct answer is A. दो भिन्न वास्तविक मूल / Two distinct real roots. Since (D=81>0), there will be two distinct real roots. (D>0) indicates different roots.
Step 3
Exam Tip
क्योंकि (D=81>0) है इसलिए दो भिन्न वास्तविक मूल होंगे। (D>0) मूलों के अलग होने का संकेत है।
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यदि किसी द्विघात समीकरण (q(x)=0) में (q(t)=0) है तो (t) क्या कहलाता है?
If (q(t)=0) in a quadratic equation (q(x)=0), what is (t) called?
#roots
#definition
#concept
A मूल / Root
B घात / Degree
C अचर पद / Constant term
D गुणांक / Coefficient
Explanation opens after your attempt
Correct Answer
A. मूल / Root
Step 1
Concept
(q(t)=0) means the equation is true when (x=t). Therefore (t) is a root of the equation.
Step 2
Why this answer is correct
The correct answer is A. मूल / Root. (q(t)=0) means the equation is true when (x=t). Therefore (t) is a root of the equation.
Step 3
Exam Tip
(q(t)=0) का अर्थ है कि (x=t) रखने पर समीकरण सत्य है। इसलिए (t) उस समीकरण का मूल है।
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यदि किसी द्विघात समीकरण के मूल (b) और (-b) हैं तो अचर पद और अग्र गुणांक के अनुपात \(\frac{c}{a}\) का मान क्या होगा?
If the roots of a quadratic equation are (b) and (-b), what is the value of the ratio \(\frac{c}{a}\) of constant term to leading coefficient?
#roots
#opposite_roots
#product
A \(-b^2\)
B \(b^2\)
C (0)
D (2b)
Explanation opens after your attempt
Correct Answer
A. \(-b^2\)
Step 1
Concept
\(\frac{c}{a}\) is the product of roots. Here (b(-b)=-b-2 ).
Step 2
Why this answer is correct
The correct answer is A. \(-b^2\). \(\frac{c}{a}\) is the product of roots. Here (b(-b)=-b-2 ).
Step 3
Exam Tip
\(\frac{c}{a}\) मूलों का गुणनफल होता है। यहां (b(-b)=-b-2 ) है।
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यदि द्विघात समीकरण के मूल (5) और \(\frac{1}{5}\) हैं तो उनके बारे में सही कथन कौन सा है?
If the roots of a quadratic equation are (5) and \(\frac{1}{5}\), which statement is correct about them?
#roots
#reciprocal_roots
#concept
A वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other
B वे बराबर मूल हैं / They are equal roots
C उनका योग (1) है / Their sum is (1)
D उनका गुणनफल (0) है / Their product is (0)
Explanation opens after your attempt
Correct Answer
A. वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other
Step 1
Concept
\(5\cdot\frac{1}{5}=1\), so the roots are reciprocals. Reciprocal roots have product (1).
Step 2
Why this answer is correct
The correct answer is A. वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other. \(5\cdot\frac{1}{5}=1\), so the roots are reciprocals. Reciprocal roots have product (1).
Step 3
Exam Tip
\(5\cdot\frac{1}{5}=1\) है इसलिए दोनों व्युत्क्रम मूल हैं। व्युत्क्रम मूलों का गुणनफल (1) होता है।
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यदि (x=2) और (x=7) किसी मोनिक द्विघात समीकरण के मूल हैं तो (x) का गुणांक क्या होगा?
If (x=2) and (x=7) are roots of a monic quadratic equation, what will be the coefficient of (x)?
#roots
#coefficient_from_roots
#monic
A (-9)
B (9)
C (14)
D (-14)
Explanation opens after your attempt
Step 1
Concept
\(The sum of roots is (2+7=9). In a monic equation the coefficient of (x) is (-(\)sum\()=-9).\)
Step 2
Why this answer is correct
\(The correct answer is A. (-9). The sum of roots is (2+7=9). In a monic equation the coefficient of (x) is (-(\)sum\()=-9).\)
Step 3
Exam Tip
मूलों का योग (2+7=9) है। मोनिक समीकरण में (x) का गुणांक (-(योग\()=-9) होता है\)।
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यदि किसी द्विघात समीकरण के मूलों का गुणनफल ऋणात्मक है तो सही कथन कौन सा है?
If the product of roots of a quadratic equation is negative, which statement is correct?
#roots
#sign_of_roots
#reasoning
A एक मूल धनात्मक और दूसरा ऋणात्मक है / One root is positive and the other is negative
B दोनों मूल धनात्मक हैं / Both roots are positive
C दोनों मूल ऋणात्मक हैं / Both roots are negative
D दोनों मूल बराबर हैं / Both roots are equal
Explanation opens after your attempt
Correct Answer
A. एक मूल धनात्मक और दूसरा ऋणात्मक है / One root is positive and the other is negative
Step 1
Concept
A negative product occurs only when the roots have opposite signs. Therefore one root will be positive and the other negative.
Step 2
Why this answer is correct
The correct answer is A. एक मूल धनात्मक और दूसरा ऋणात्मक है / One root is positive and the other is negative. A negative product occurs only when the roots have opposite signs. Therefore one root will be positive and the other negative.
Step 3
Exam Tip
ऋणात्मक गुणनफल तभी होता है जब मूलों के चिन्ह विपरीत हों। इसलिए एक मूल धनात्मक और दूसरा ऋणात्मक होगा।
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यदि किसी द्विघात समीकरण में (a=1), (b=-16), (c=64) है तो मूल कौन से होंगे?
If a quadratic equation has (a=1), (b=-16), and (c=64), what will be the roots?
#roots
#coefficients
#repeated_root
A (8) और (8) / (8) and (8)
B (-8) और (-8) / (-8) and (-8)
C (16) और (64) / (16) and (64)
D (0) और (8) / (0) and (8)
Explanation opens after your attempt
Correct Answer
A. (8) और (8) / (8) and (8)
Step 1
Concept
The equation is \(x^2-16x+64=0\), which becomes ((x-8)2 =0). Therefore both roots are (8).
Step 2
Why this answer is correct
The correct answer is A. (8) और (8) / (8) and (8). The equation is \(x^2-16x+64=0\), which becomes ((x-8)2 =0). Therefore both roots are (8).
Step 3
Exam Tip
समीकरण \(x^2-16x+64=0\) है जो ((x-8)2 =0) बनता है। इसलिए दोनों मूल (8) हैं।
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यदि (x-4) और (x+7) किसी द्विघात समीकरण के गुणनखंड हैं तो उसके मूल कौन से हैं?
If (x-4) and (x+7) are factors of a quadratic equation, what are its roots?
#roots
#factors_to_roots
#zero_product
A (4) और (-7) / (4) and (-7)
B (-4) और (7) / (-4) and (7)
C (4) और (7) / (4) and (7)
D (-4) और (-7) / (-4) and (-7)
Explanation opens after your attempt
Correct Answer
A. (4) और (-7) / (4) and (-7)
Step 1
Concept
From (x-4=0), (x=4), and from (x+7=0), (x=-7). The sign changes when finding a root from a factor.
Step 2
Why this answer is correct
The correct answer is A. (4) और (-7) / (4) and (-7). From (x-4=0), (x=4), and from (x+7=0), (x=-7). The sign changes when finding a root from a factor.
Step 3
Exam Tip
(x-4=0) से (x=4) और (x+7=0) से (x=-7) मिलता है। गुणनखंड का चिन्ह उलटकर मूल मिलता है।
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यदि द्विघात समीकरण के मूल (2) और (-7) हैं तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) का मान क्या होगा?
If the roots of a quadratic equation are (2) and (-7), what is the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\)?
#roots
#reciprocal_sum
#identity
A \(\frac{5}{14}\)
B -\(\frac{5}{14}\)
C \(\frac{9}{14}\)
D -\(\frac{9}{14}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{5}{14}\)
Step 1
Concept
Here \(\alpha+\beta=-5\) and \(\alpha\beta=-14\). Therefore \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-14}=\frac{5}{14}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{5}{14}\). Here \(\alpha+\beta=-5\) and \(\alpha\beta=-14\). Therefore \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-14}=\frac{5}{14}\).
Step 3
Exam Tip
यहां \(\alpha+\beta=-5\) और \(\alpha\beta=-14\) है। इसलिए \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{-5}{-14}=\frac{5}{14}\) है।
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यदि मूलों का योग (9) और गुणनफल (18) है तो मोनिक द्विघात समीकरण कौन सा होगा?
If the sum of roots is (9) and product is (18), which monic quadratic equation is formed?
#roots
#equation_from_sum_product
#monic
A \(x^2-9x+18=0\)
B \(x^2+9x+18=0\)
C \(x^2-18x+9=0\)
D \(x^2+18x+9=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-9x+18=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-9x+18=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-9x+18=0). A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-9x+18=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2-9x+18=0) सही है\)।
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यदि किसी द्विघात समीकरण के लिए (D=64) है तो उसके वास्तविक मूलों की प्रकृति क्या होगी?
If (D=64) for a quadratic equation, what will be the nature of its real roots?
#roots
#discriminant
#nature
A दो भिन्न वास्तविक मूल / Two distinct real roots
B दो बराबर वास्तविक मूल / Two equal real roots
C कोई वास्तविक मूल नहीं / No real root
D एक ही मूल (0) / Only one root (0)
Explanation opens after your attempt
Correct Answer
A. दो भिन्न वास्तविक मूल / Two distinct real roots
Step 1
Concept
Since (D=64>0), there will be two distinct real roots. When (D>0), the roots are different.
Step 2
Why this answer is correct
The correct answer is A. दो भिन्न वास्तविक मूल / Two distinct real roots. Since (D=64>0), there will be two distinct real roots. When (D>0), the roots are different.
Step 3
Exam Tip
क्योंकि (D=64>0) है इसलिए दो भिन्न वास्तविक मूल होंगे। (D>0) होने पर मूल अलग अलग होते हैं।
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यदि किसी द्विघात समीकरण (p(x)=0) में (x=a) रखने पर (p(a)=0) हो जाता है तो (a) क्या कहलाता है?
If substituting (x=a) in a quadratic equation (p(x)=0) gives (p(a)=0), what is (a) called?
#roots
#definition
#substitution
A मूल / Root
B अचर पद / Constant term
C मध्य पद / Middle term
D अग्र गुणांक / Leading coefficient
Explanation opens after your attempt
Correct Answer
A. मूल / Root
Step 1
Concept
Since the equation becomes true after substituting (x=a), (a) is a root. In exams check a root by direct substitution.
Step 2
Why this answer is correct
The correct answer is A. मूल / Root. Since the equation becomes true after substituting (x=a), (a) is a root. In exams check a root by direct substitution.
Step 3
Exam Tip
क्योंकि (x=a) रखने पर समीकरण सत्य हो जाता है इसलिए (a) मूल है। परीक्षा में मूल की जांच सीधे प्रतिस्थापन से करें।
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यदि (D<0) है तो किसी द्विघात समीकरण के वास्तविक मूलों की संख्या कितनी होगी?
If (D<0), how many real roots will a quadratic equation have?
#roots
#discriminant
#no_real_roots
A (0)
B (1)
C (2)
D अनगिनत / Infinitely many
Explanation opens after your attempt
Step 1
Concept
When (D<0), there are no real roots. Therefore the number of real roots is (0).
Step 2
Why this answer is correct
The correct answer is A. (0). When (D<0), there are no real roots. Therefore the number of real roots is (0).
Step 3
Exam Tip
(D<0) होने पर वास्तविक मूल नहीं होते। इसलिए वास्तविक मूलों की संख्या (0) होगी।
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यदि किसी द्विघात समीकरण के मूल (a) और (-a) हैं तो अचर पद और अग्र गुणांक के अनुपात \(\frac{c}{a_1}\) का मान क्या होगा?
If the roots of a quadratic equation are (a) and (-a), what is the value of the ratio \(\frac{c}{a_1}\) of constant term to leading coefficient?
#roots
#opposite_roots
#product
A \(-a^2\)
B \(a^2\)
C (0)
D (2a)
Explanation opens after your attempt
Correct Answer
A. \(-a^2\)
Step 1
Concept
\(\frac{c}{a_1}\) is the product of roots. Here (a(-a)=-a-2 ).
Step 2
Why this answer is correct
The correct answer is A. \(-a^2\). \(\frac{c}{a_1}\) is the product of roots. Here (a(-a)=-a-2 ).
Step 3
Exam Tip
\(\frac{c}{a_1}\) मूलों का गुणनफल होता है। यहां (a(-a)=-a-2 ) है।
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यदि द्विघात समीकरण के मूल (4) और \(\frac{1}{4}\) हैं तो उनके बारे में सही कथन कौन सा है?
If the roots of a quadratic equation are (4) and \(\frac{1}{4}\), which statement is correct about them?
#roots
#reciprocal_roots
#concept
A वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other
B वे बराबर मूल हैं / They are equal roots
C उनका योग (1) है / Their sum is (1)
D उनका गुणनफल (0) है / Their product is (0)
Explanation opens after your attempt
Correct Answer
A. वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other
Step 1
Concept
\(4\cdot\frac{1}{4}=1\), so the roots are reciprocals. Reciprocal roots have product (1).
Step 2
Why this answer is correct
The correct answer is A. वे एक दूसरे के व्युत्क्रम हैं / They are reciprocals of each other. \(4\cdot\frac{1}{4}=1\), so the roots are reciprocals. Reciprocal roots have product (1).
Step 3
Exam Tip
\(4\cdot\frac{1}{4}=1\) है इसलिए दोनों व्युत्क्रम मूल हैं। व्युत्क्रम मूलों का गुणनफल (1) होता है।
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यदि (x=1) और (x=4) किसी मोनिक द्विघात समीकरण के मूल हैं तो (x) का गुणांक क्या होगा?
If (x=1) and (x=4) are roots of a monic quadratic equation, what will be the coefficient of (x)?
#roots
#coefficient_from_roots
#monic
A (-5)
B (5)
C (4)
D (-4)
Explanation opens after your attempt
Step 1
Concept
\(The sum of roots is (1+4=5). In a monic equation the coefficient of (x) is (-(\)sum\()=-5).\)
Step 2
Why this answer is correct
\(The correct answer is A. (-5). The sum of roots is (1+4=5). In a monic equation the coefficient of (x) is (-(\)sum\()=-5).\)
Step 3
Exam Tip
मूलों का योग (1+4=5) है। मोनिक समीकरण में (x) का गुणांक (-(योग\()=-5) होता है\)।
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यदि किसी द्विघात समीकरण के मूलों का गुणनफल (0) है तो सही कथन कौन सा है?
If the product of roots of a quadratic equation is (0), which statement is correct?
#roots
#zero_product
#reasoning
A कम से कम एक मूल (0) है / At least one root is (0)
B दोनों मूल जरूर (1) हैं / Both roots must be (1)
C दोनों मूल ऋणात्मक हैं / Both roots are negative
D कोई वास्तविक मूल नहीं है / There is no real root
Explanation opens after your attempt
Correct Answer
A. कम से कम एक मूल (0) है / At least one root is (0)
Step 1
Concept
If \(\alpha\beta=0\), then \(\alpha=0\) or \(\beta=0\). If the product is zero, always check for a zero root.
Step 2
Why this answer is correct
The correct answer is A. कम से कम एक मूल (0) है / At least one root is (0). If \(\alpha\beta=0\), then \(\alpha=0\) or \(\beta=0\). If the product is zero, always check for a zero root.
Step 3
Exam Tip
यदि \(\alpha\beta=0\) है तो \(\alpha=0\) या \(\beta=0\) होगा। गुणनफल शून्य हो तो शून्य मूल जरूर देखें।
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यदि किसी द्विघात समीकरण में (a=1), (b=-10), (c=25) है तो मूल कौन से होंगे?
If a quadratic equation has (a=1), (b=-10), and (c=25), what will be the roots?
#roots
#coefficients
#repeated_root
A (5) और (5) / (5) and (5)
B (-5) और (-5) / (-5) and (-5)
C (10) और (25) / (10) and (25)
D (0) और (5) / (0) and (5)
Explanation opens after your attempt
Correct Answer
A. (5) और (5) / (5) and (5)
Step 1
Concept
The equation is \(x^2-10x+25=0\), which becomes ((x-5)2 =0). Therefore both roots are (5).
Step 2
Why this answer is correct
The correct answer is A. (5) और (5) / (5) and (5). The equation is \(x^2-10x+25=0\), which becomes ((x-5)2 =0). Therefore both roots are (5).
Step 3
Exam Tip
समीकरण \(x^2-10x+25=0\) है जो ((x-5)2 =0) बनता है। इसलिए दोनों मूल (5) हैं।
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यदि (x+2) और (x-5) किसी द्विघात समीकरण के गुणनखंड हैं तो उसके मूल कौन से हैं?
If (x+2) and (x-5) are factors of a quadratic equation, what are its roots?
#roots
#factors_to_roots
#zero_product
A (-2) और (5) / (-2) and (5)
B (2) और (-5) / (2) and (-5)
C (2) और (5) / (2) and (5)
D (-2) और (-5) / (-2) and (-5)
Explanation opens after your attempt
Correct Answer
A. (-2) और (5) / (-2) and (5)
Step 1
Concept
From (x+2=0), (x=-2), and from (x-5=0), (x=5). The sign changes when finding a root from a factor.
Step 2
Why this answer is correct
The correct answer is A. (-2) और (5) / (-2) and (5). From (x+2=0), (x=-2), and from (x-5=0), (x=5). The sign changes when finding a root from a factor.
Step 3
Exam Tip
(x+2=0) से (x=-2) और (x-5=0) से (x=5) मिलता है। गुणनखंड का चिन्ह उलटकर मूल मिलता है।
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यदि द्विघात समीकरण के मूल (3) और (-4) हैं तो \(\frac{1}{\alpha}+\frac{1}{\beta}\) का मान क्या होगा?
If the roots of a quadratic equation are (3) and (-4), what is the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\)?
#roots
#reciprocal_sum
#identity
A -\(\frac{1}{12}\)
B \(\frac{1}{12}\)
C -\(\frac{7}{12}\)
D \(\frac{7}{12}\)
Explanation opens after your attempt
Correct Answer
A. -\(\frac{1}{12}\)
Step 1
Concept
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here the sum is (-1) and product is (-12), so the value is \(\frac{1}{12}\).
Step 2
Why this answer is correct
The correct answer is A. -\(\frac{1}{12}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\). Here the sum is (-1) and product is (-12), so the value is \(\frac{1}{12}\).
Step 3
Exam Tip
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}\) है। यहां \(\frac{-1}{-12}\) नहीं बल्कि \(-\frac{1}{12}\) क्योंकि योग (-1) और गुणनफल (-12) है।
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यदि मूलों का योग (7) और गुणनफल (10) है तो मोनिक द्विघात समीकरण कौन सा होगा?
If the sum of roots is (7) and product is (10), which monic quadratic equation is formed?
#roots
#equation_from_sum_product
#monic
A \(x^2-7x+10=0\)
B \(x^2+7x+10=0\)
C \(x^2-10x+7=0\)
D \(x^2+10x+7=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-7x+10=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-7x+10=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-7x+10=0). A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-7x+10=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2-7x+10=0) सही है\)।
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यदि किसी द्विघात समीकरण के लिए (D=49) है तो उसके वास्तविक मूलों की प्रकृति क्या होगी?
If (D=49) for a quadratic equation, what will be the nature of its real roots?
#roots
#discriminant
#nature
A दो भिन्न वास्तविक मूल / Two distinct real roots
B दो बराबर वास्तविक मूल / Two equal real roots
C कोई वास्तविक मूल नहीं / No real root
D केवल एक मूल (0) / Only one root (0)
Explanation opens after your attempt
Correct Answer
A. दो भिन्न वास्तविक मूल / Two distinct real roots
Step 1
Concept
Since (D=49>0), two distinct real roots will be obtained. When (D>0), the roots are different.
Step 2
Why this answer is correct
The correct answer is A. दो भिन्न वास्तविक मूल / Two distinct real roots. Since (D=49>0), two distinct real roots will be obtained. When (D>0), the roots are different.
Step 3
Exam Tip
क्योंकि (D=49>0) है इसलिए दो भिन्न वास्तविक मूल मिलेंगे। (D>0) होने पर मूल अलग अलग होते हैं।
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यदि किसी द्विघात समीकरण के मूल (r) और (r) हैं तो उनके गुणनफल का मान क्या होगा?
If the roots of a quadratic equation are (r) and (r), what will be their product?
#roots
#equal_roots
#product
A (2r)
B (0)
C \(r^2\)
D (r)
Explanation opens after your attempt
Correct Answer
C. \(r^2\)
Step 1
Concept
The product of (r) and (r) is \(r^2\). For equal roots, remember sum (2r) and product \(r^2\).
Step 2
Why this answer is correct
The correct answer is C. \(r^2\). The product of (r) and (r) is \(r^2\). For equal roots, remember sum (2r) and product \(r^2\).
Step 3
Exam Tip
(r) और (r) का गुणनफल \(r^2\) होता है। बराबर मूलों में योग (2r) और गुणनफल \(r^2\) याद रखें।
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जिस मोनिक द्विघात समीकरण के मूलों का योग (10) और गुणनफल (21) है वह कौन सा है?
Which monic quadratic equation has sum of roots (10) and product of roots (21)?
#roots
#equation_from_sum_product
#monic
A \(x^2+10x+21=0\)
B \(x^2-10x+21=0\)
C \(x^2-21x+10=0\)
D \(x^2+21x+10=0\)
Explanation opens after your attempt
Correct Answer
B. \(x^2-10x+21=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-10x+21=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is B. (x^2-10x+21=0). A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-10x+21=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2-10x+21=0) सही है\)।
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किस स्थिति में द्विघात समीकरण के दो भिन्न वास्तविक मूल होते हैं?
In which condition does a quadratic equation have two distinct real roots?
#roots
#discriminant
#distinct_roots
A (D=0)
B (D<0)
C (D>0)
D (a=0)
Explanation opens after your attempt
Step 1
Concept
For two distinct real roots, (D>0) is required. This is a direct rule to check the nature.
Step 2
Why this answer is correct
The correct answer is C. (D>0). For two distinct real roots, (D>0) is required. This is a direct rule to check the nature.
Step 3
Exam Tip
दो भिन्न वास्तविक मूलों के लिए (D>0) होना चाहिए। यह प्रकृति जांचने का सीधा नियम है।
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यदि (D=0) है तो द्विघात समीकरण के वास्तविक मूल कैसे होते हैं?
If (D=0), how are the real roots of a quadratic equation?
#roots
#discriminant
#equal_roots
A दो भिन्न वास्तविक मूल / Two distinct real roots
B दो बराबर वास्तविक मूल / Two equal real roots
C कोई वास्तविक मूल नहीं / No real root
D चार वास्तविक मूल / Four real roots
Explanation opens after your attempt
Correct Answer
B. दो बराबर वास्तविक मूल / Two equal real roots
Step 1
Concept
When (D=0), the two real roots are equal. Check \(D=b^2-4ac\) for the nature of roots.
Step 2
Why this answer is correct
The correct answer is B. दो बराबर वास्तविक मूल / Two equal real roots. When (D=0), the two real roots are equal. Check \(D=b^2-4ac\) for the nature of roots.
Step 3
Exam Tip
(D=0) होने पर दोनों वास्तविक मूल बराबर होते हैं। मूलों की प्रकृति के लिए \(D=b^2-4ac\) देखें।
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यदि किसी द्विघात समीकरण के मूल (r) और (-r) हैं तो उनके योग का मान क्या होगा?
If the roots of a quadratic equation are (r) and (-r), what is their sum?
#roots
#opposite_roots
#sum
A (r)
B (2r)
C (0)
D \(r^2\)
Explanation opens after your attempt
Step 1
Concept
(r+(-r)=0). The sum of opposite roots is always (0).
Step 2
Why this answer is correct
The correct answer is C. (0). (r+(-r)=0). The sum of opposite roots is always (0).
Step 3
Exam Tip
(r+(-r)=0) होता है। विपरीत मूलों का योग हमेशा (0) होता है।
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जिस मोनिक द्विघात समीकरण के मूलों का योग (-9) और गुणनफल (20) है वह कौन सा है?
Which monic quadratic equation has sum of roots (-9) and product of roots (20)?
#roots
#equation_from_sum_product
#monic
A \(x^2+9x+20=0\)
B \(x^2-9x+20=0\)
C \(x^2+20x+9=0\)
D \(x^2-20x+9=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+9x+20=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2+9x+20=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2+9x+20=0). A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2+9x+20=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2+9x+20=0) सही है\)।
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किस स्थिति में द्विघात समीकरण के कोई वास्तविक मूल नहीं होते?
In which condition does a quadratic equation have no real roots?
#roots
#discriminant
#condition
A (D>0)
B (D=0)
C (D<0)
D \(D\ge0\)
Explanation opens after your attempt
Step 1
Concept
When (D<0), there are no real roots. This is a direct rule for the nature of roots.
Step 2
Why this answer is correct
The correct answer is C. (D<0). When (D<0), there are no real roots. This is a direct rule for the nature of roots.
Step 3
Exam Tip
जब (D<0) होता है तब वास्तविक मूल नहीं होते। यह मूलों की प्रकृति का सीधा नियम है।
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यदि (D>0) है तो द्विघात समीकरण के वास्तविक मूल कैसे होते हैं?
If (D>0), how are the real roots of a quadratic equation?
#roots
#discriminant
#nature
A दो बराबर वास्तविक मूल / Two equal real roots
B दो भिन्न वास्तविक मूल / Two distinct real roots
C कोई वास्तविक मूल नहीं / No real root
D केवल एक अचर पद / Only one constant term
Explanation opens after your attempt
Correct Answer
B. दो भिन्न वास्तविक मूल / Two distinct real roots
Step 1
Concept
When (D>0), two distinct real roots are obtained. To know the nature of roots, check \(D=b^2-4ac\).
Step 2
Why this answer is correct
The correct answer is B. दो भिन्न वास्तविक मूल / Two distinct real roots. When (D>0), two distinct real roots are obtained. To know the nature of roots, check \(D=b^2-4ac\).
Step 3
Exam Tip
(D>0) होने पर दो भिन्न वास्तविक मूल मिलते हैं। मूलों की प्रकृति जानने के लिए \(D=b^2-4ac\) देखें।
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किसी द्विघात समीकरण (p(x)=0) का मूल वह संख्या है जिससे (p(x)) का मान क्या हो जाता है?
A root of a quadratic equation (p(x)=0) is a number for which the value of (p(x)) becomes what?
#roots
#definition
#concept
A (1)
B (0)
C (-1)
D (p)
Explanation opens after your attempt
Step 1
Concept
When a root is substituted, (p(x)=0) becomes true. In exams always check a root by substitution.
Step 2
Why this answer is correct
The correct answer is B. (0). When a root is substituted, (p(x)=0) becomes true. In exams always check a root by substitution.
Step 3
Exam Tip
मूल रखने पर (p(x)=0) बनता है। परीक्षा में मूल की जांच हमेशा प्रतिस्थापन से करें।
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यदि किसी द्विघात समीकरण के मूल एक दूसरे के व्युत्क्रम हैं तो उनके गुणनफल का मान क्या होगा?
If the roots of a quadratic equation are reciprocals of each other then what is their product?
#roots
#reciprocal_roots
#product
A (0)
B (1)
C (-1)
D (2)
Explanation opens after your attempt
Step 1
Concept
If the roots are (r) and \(\frac{1}{r}\) their product is (1). For reciprocal roots remember the product is (1).
Step 2
Why this answer is correct
The correct answer is B. (1). If the roots are (r) and \(\frac{1}{r}\) their product is (1). For reciprocal roots remember the product is (1).
Step 3
Exam Tip
यदि मूल (r) और \(\frac{1}{r}\) हों तो गुणनफल (1) होता है। व्युत्क्रम मूलों में गुणनफल तुरंत (1) याद रखें।
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जिस द्विघात समीकरण के मूलों का योग (6) और गुणनफल (8) है वह कौन सा है?
Which quadratic equation has sum of roots (6) and product of roots (8)?
#roots
#equation_from_sum_product
#formula
A \(x^2+6x+8=0\)
B \(x^2-6x+8=0\)
C \(x^2-8x+6=0\)
D \(x^2+8x+6=0\)
Explanation opens after your attempt
Correct Answer
B. \(x^2-6x+8=0\)
Step 1
Concept
\(The standard form is (x^2-(\)sum)x+product\(=0) so (x^2-6x+8=0). The sign of the sum term changes.\)
Step 2
Why this answer is correct
\(The correct answer is B. (x^2-6x+8=0). The standard form is (x^2-(\)sum)x+product\(=0) so (x^2-6x+8=0). The sign of the sum term changes.\)
Step 3
Exam Tip
\(मानक रूप (x^2-(\)योग)x+गुणनफल=0) है इसलिए \(x^2-6x+8=0\)। योग वाले पद का चिन्ह बदलता है।
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यदि किसी द्विघात समीकरण के वास्तविक मूलों का गुणनफल ऋणात्मक है तो मूलों के चिन्ह कैसे होंगे?
If the product of real roots of a quadratic equation is negative then how are the signs of the roots?
#roots
#sign_of_roots
#reasoning
A दोनों धनात्मक / Both positive
B दोनों ऋणात्मक / Both negative
C एक धनात्मक और एक ऋणात्मक / One positive and one negative
D दोनों शून्य / Both zero
Explanation opens after your attempt
Correct Answer
C. एक धनात्मक और एक ऋणात्मक / One positive and one negative
Step 1
Concept
A negative product occurs when one root is positive and the other is negative. \(\alpha\beta<0\) is a quick sign check.
Step 2
Why this answer is correct
The correct answer is C. एक धनात्मक और एक ऋणात्मक / One positive and one negative. A negative product occurs when one root is positive and the other is negative. \(\alpha\beta<0\) is a quick sign check.
Step 3
Exam Tip
ऋणात्मक गुणनफल तभी मिलता है जब एक मूल धनात्मक और दूसरा ऋणात्मक हो। \(\alpha\beta<0\) संकेतों की जांच का छोटा संकेत है।
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किस स्थिति में द्विघात समीकरण के वास्तविक मूल होते हैं?
In which condition does a quadratic equation have real roots?
#roots
#real_roots
#discriminant
A (D<0)
B \(D\ge 0\)
C (D=-1)
D (a=0)
Explanation opens after your attempt
Correct Answer
B. \(D\ge 0\)
Step 1
Concept
For real roots \(D\ge 0\) is required. If (D<0) there are no real roots.
Step 2
Why this answer is correct
The correct answer is B. \(D\ge 0\). For real roots \(D\ge 0\) is required. If (D<0) there are no real roots.
Step 3
Exam Tip
वास्तविक मूलों के लिए \(D\ge 0\) होना चाहिए। यदि (D<0) हो तो वास्तविक मूल नहीं मिलते।
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यदि किसी द्विघात समीकरण के दो बराबर वास्तविक मूल हों तो डिस्क्रिमिनेंट (D) का मान कैसा होता है?
If a quadratic equation has two equal real roots then what is the value of discriminant (D)?
#roots
#discriminant
#equal_roots
A (D>0)
B (D=0)
C (D<0)
D (D=1)
Explanation opens after your attempt
Step 1
Concept
For equal real roots (D=0). The discriminant quickly tells the nature of roots.
Step 2
Why this answer is correct
The correct answer is B. (D=0). For equal real roots (D=0). The discriminant quickly tells the nature of roots.
Step 3
Exam Tip
बराबर वास्तविक मूलों के लिए (D=0) होता है। डिस्क्रिमिनेंट से मूलों की प्रकृति जल्दी पता चलती है।
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किसी द्विघात समीकरण का मूल किसे कहते हैं?
What is called a root of a quadratic equation?
#roots
#definition
#concept
A वह मान जिससे समीकरण सत्य हो / The value that satisfies the equation
B केवल सबसे बड़ा गुणांक / Only the greatest coefficient
C केवल अचर पद / Only the constant term
D समीकरण की घात / The degree of the equation
Explanation opens after your attempt
Correct Answer
A. वह मान जिससे समीकरण सत्य हो / The value that satisfies the equation
Step 1
Concept
A root is a value that makes the equation equal to (0). In exams first substitute the value and check.
Step 2
Why this answer is correct
The correct answer is A. वह मान जिससे समीकरण सत्य हो / The value that satisfies the equation. A root is a value that makes the equation equal to (0). In exams first substitute the value and check.
Step 3
Exam Tip
मूल वह मान है जिसे रखने पर समीकरण का मान (0) हो जाता है। परीक्षा में पहले मान रखकर जांचें।
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यदि \(x^2-7x+10=0\) के मूलों को उलटकर नया समीकरण बनाया जाए तो नया मोनिक समीकरण कौन सा होगा?
If a new equation is formed by taking reciprocals of the roots of \(x^2-7x+10=0\), which monic equation is obtained?
#roots
#reciprocal_roots
#new_equation
A \(x^2-\frac{7}{10}x+\frac{1}{10}=0\)
B \(x^2-7x+10=0\)
C \(x^2+\frac{7}{10}x+\frac{1}{10}=0\)
D \(x^2-\frac{1}{10}x+\frac{7}{10}=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-\frac{7}{10}x+\frac{1}{10}=0\)
Step 1
Concept
The old sum is (7) and product is (10). The reciprocal roots have sum \(\frac{7}{10}\) and product \(\frac{1}{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-\frac{7}{10}x+\frac{1}{10}=0\). The old sum is (7) and product is (10). The reciprocal roots have sum \(\frac{7}{10}\) and product \(\frac{1}{10}\).
Step 3
Exam Tip
पुराने योग (7) और गुणनफल (10) हैं। उलटे मूलों का योग \(\frac{7}{10}\) और गुणनफल \(\frac{1}{10}\) होगा।
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यदि \(x^2-4x+3=0\) के मूल \(\alpha\) और \(\beta\) हैं तो \(3\alpha\) और \(3\beta\) को मूल मानकर समीकरण कौन सा होगा?
If \(\alpha\) and \(\beta\) are roots of \(x^2-4x+3=0\), which equation has \(3\alpha\) and \(3\beta\) as roots?
#roots
#transformed_roots
#equation
A \(x^2-12x+27=0\)
B \(x^2-4x+27=0\)
C \(x^2-12x+9=0\)
D \(x^2+12x+27=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-12x+27=0\)
Step 1
Concept
The old sum is (4) and product is (3). The new sum is (12) and product is (27), so the equation is \(x^2-12x+27=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-12x+27=0\). The old sum is (4) and product is (3). The new sum is (12) and product is (27), so the equation is \(x^2-12x+27=0\).
Step 3
Exam Tip
पुराने योग (4) और गुणनफल (3) हैं। नए योग (12) और गुणनफल (27) होंगे इसलिए \(x^2-12x+27=0\) है।
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यदि \(x^2-5x+6=0\) के मूलों को उलटकर नया समीकरण बनाया जाए तो नया मोनिक समीकरण कौन सा होगा?
If a new equation is formed by taking reciprocals of the roots of \(x^2-5x+6=0\), which monic equation is obtained?
#roots
#reciprocal_roots
#new_equation
A \(x^2-\frac{5}{6}x+\frac{1}{6}=0\)
B \(x^2-5x+6=0\)
C \(x^2+\frac{5}{6}x+\frac{1}{6}=0\)
D \(x^2-\frac{1}{6}x+\frac{5}{6}=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-\frac{5}{6}x+\frac{1}{6}=0\)
Step 1
Concept
The old sum is (5) and product is (6). The reciprocal roots have sum \(\frac{5}{6}\) and product \(\frac{1}{6}\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-\frac{5}{6}x+\frac{1}{6}=0\). The old sum is (5) and product is (6). The reciprocal roots have sum \(\frac{5}{6}\) and product \(\frac{1}{6}\).
Step 3
Exam Tip
पुराने योग (5) और गुणनफल (6) हैं। उलटे मूलों का योग \(\frac{5}{6}\) और गुणनफल \(\frac{1}{6}\) होगा।
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यदि \(x^2-3x+2=0\) के मूल \(\alpha\) और \(\beta\) हैं तो \(2\alpha\) और \(2\beta\) को मूल मानकर समीकरण कौन सा होगा?
If \(\alpha\) and \(\beta\) are roots of \(x^2-3x+2=0\), which equation has \(2\alpha\) and \(2\beta\) as roots?
#roots
#transformed_roots
#equation
A \(x^2-6x+8=0\)
B \(x^2-3x+8=0\)
C \(x^2-6x+4=0\)
D \(x^2+6x+8=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x+8=0\)
Step 1
Concept
The old sum is (3) and product is (2). The new sum is (6) and product is (8), so the equation is \(x^2-6x+8=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x+8=0\). The old sum is (3) and product is (2). The new sum is (6) and product is (8), so the equation is \(x^2-6x+8=0\).
Step 3
Exam Tip
पुराने योग (3) और गुणनफल (2) हैं। नए योग (6) और गुणनफल (8) होंगे इसलिए \(x^2-6x+8=0\) है।
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किस द्विघात समीकरण के मूलों का योग (0) और गुणनफल (-169) है?
Which quadratic equation has sum of roots (0) and product (-169)?
#quadratic-equations
#sum-product
#forming-equation
#expert
A \(x^2-169=0\)
B \(x^2+169=0\)
C \(x^2+13x-169=0\)
D \(x^2-13x-169=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-169=0\)
Step 1
Concept
\(The monic equation is (x^2-(\)sum)x+product\(=0). Using sum (0) and product (-169) gives (x^2-169=0).\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-169=0). The monic equation is (x^2-(\)sum)x+product\(=0). Using sum (0) and product (-169) gives (x^2-169=0).\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) है। \(योग (0) और गुणनफल (-169) रखने पर (x^2-169=0) मिलता है\)।
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दो संख्याओं का योग (19) और गुणनफल (90) है। वे किस द्विघात समीकरण के मूल हो सकते हैं?
Two numbers have sum (19) and product (90). They can be roots of which quadratic equation?
#quadratic-equations
#forming-equation
#sum-product
#expert
A \(x^2-19x+90=0\)
B \(x^2+19x+90=0\)
C \(x^2-90x+19=0\)
D \(x^2+90x-19=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-19x+90=0\)
Step 1
Concept
If the sum of roots is (19) and product is (90), the equation is \(x^2-19x+90=0\). Remember the monic form formula.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-19x+90=0\). If the sum of roots is (19) and product is (90), the equation is \(x^2-19x+90=0\). Remember the monic form formula.
Step 3
Exam Tip
यदि मूलों का योग (19) और गुणनफल (90) है, तो समीकरण \(x^2-19x+90=0\) होगा। मोनिक रूप का सूत्र याद रखें।
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यदि किसी मोनिक द्विघात समीकरण के मूल (3r) और (4r) हैं तथा उनका योग (28) है, तो उस समीकरण का स्थिर पद क्या होगा?
If the roots of a monic quadratic equation are (3r) and (4r), and their sum is (28), what will be the constant term?
#quadratic-equations
#roots
#monic-equation
#expert
A (192)
B (28)
C (112)
D (84)
Explanation opens after your attempt
Step 1
Concept
From (3r+4r=28), we get (r=4), so the roots are (12) and (16). The constant term is the product of roots (192).
Step 2
Why this answer is correct
The correct answer is A. (192). From (3r+4r=28), we get (r=4), so the roots are (12) and (16). The constant term is the product of roots (192).
Step 3
Exam Tip
(3r+4r=28) से (r=4) मिलता है, इसलिए मूल (12) और (16) हैं। स्थिर पद मूलों का गुणनफल (192) होगा।
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मूलों का योग (-15) और गुणनफल (56) वाला मोनिक द्विघात समीकरण कौन-सा है?
Which monic quadratic equation has sum of roots (-15) and product (56)?
#quadratic-equations
#sum-product
#forming-equation
#expert
A \(x^2+15x+56=0\)
B \(x^2-15x+56=0\)
C \(x^2+56x+15=0\)
D \(x^2-56x+15=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+15x+56=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Substituting sum (-15) gives (x^2+15x+56=0).\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2+15x+56=0). A monic equation is (x^2-(\)sum)x+product\(=0). Substituting sum (-15) gives (x^2+15x+56=0).\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(योग (-15) रखने पर (x^2+15x+56=0) मिलता है\)।
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किस द्विघात समीकरण के मूलों का योग (0) और गुणनफल (-144) है?
Which quadratic equation has sum of roots (0) and product (-144)?
#quadratic-equations
#sum-product
#forming-equation
#expert
A \(x^2-144=0\)
B \(x^2+144=0\)
C \(x^2+12x-144=0\)
D \(x^2-12x-144=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-144=0\)
Step 1
Concept
\(The monic equation is (x^2-(\)sum)x+product\(=0). Using sum (0) and product (-144) gives (x^2-144=0).\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-144=0). The monic equation is (x^2-(\)sum)x+product\(=0). Using sum (0) and product (-144) gives (x^2-144=0).\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) है। \(योग (0) और गुणनफल (-144) रखने पर (x^2-144=0) मिलता है\)।
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दो संख्याओं का योग (17) और गुणनफल (72) है। वे किस द्विघात समीकरण के मूल हो सकते हैं?
Two numbers have sum (17) and product (72). They can be roots of which quadratic equation?
#quadratic-equations
#forming-equation
#sum-product
#expert
A \(x^2-17x+72=0\)
B \(x^2+17x+72=0\)
C \(x^2-72x+17=0\)
D \(x^2+72x-17=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-17x+72=0\)
Step 1
Concept
If the sum of roots is (17) and product is (72), the equation is \(x^2-17x+72=0\). Remember the monic form formula.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-17x+72=0\). If the sum of roots is (17) and product is (72), the equation is \(x^2-17x+72=0\). Remember the monic form formula.
Step 3
Exam Tip
यदि मूलों का योग (17) और गुणनफल (72) है, तो समीकरण \(x^2-17x+72=0\) होगा। मोनिक रूप का सूत्र याद रखें।
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यदि किसी मोनिक द्विघात समीकरण के मूल (2r) और (5r) हैं तथा उनका योग (21) है, तो उस समीकरण का स्थिर पद क्या होगा?
If the roots of a monic quadratic equation are (2r) and (5r), and their sum is (21), what will be the constant term?
#quadratic-equations
#roots
#monic-equation
#expert
A (90)
B (21)
C (45)
D (63)
Explanation opens after your attempt
Step 1
Concept
From (2r+5r=21), we get (r=3), so the roots are (6) and (15). The constant term is the product of roots (90).
Step 2
Why this answer is correct
The correct answer is A. (90). From (2r+5r=21), we get (r=3), so the roots are (6) and (15). The constant term is the product of roots (90).
Step 3
Exam Tip
(2r+5r=21) से (r=3) मिलता है, इसलिए मूल (6) और (15) हैं। स्थिर पद मूलों का गुणनफल (90) होगा।
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मूलों का योग (-13) और गुणनफल (42) वाला मोनिक द्विघात समीकरण कौन-सा है?
Which monic quadratic equation has sum of roots (-13) and product (42)?
#quadratic-equations
#sum-product
#forming-equation
#expert
A \(x^2+13x+42=0\)
B \(x^2-13x+42=0\)
C \(x^2+42x+13=0\)
D \(x^2-42x+13=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+13x+42=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Substituting sum (-13) gives (x^2+13x+42=0).\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2+13x+42=0). A monic equation is (x^2-(\)sum)x+product\(=0). Substituting sum (-13) gives (x^2+13x+42=0).\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(योग (-13) रखने पर (x^2+13x+42=0) मिलता है\)।
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किस द्विघात समीकरण के मूलों का योग (0) और गुणनफल (-121) है?
Which quadratic equation has sum of roots (0) and product (-121)?
#quadratic-equations
#sum-product
#forming-equation
#expert
A \(x^2-121=0\)
B \(x^2+121=0\)
C \(x^2+11x-121=0\)
D \(x^2-11x-121=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-121=0\)
Step 1
Concept
\(The monic equation is (x^2-(\)sum)x+product\(=0). Using sum (0) and product (-121) gives (x^2-121=0).\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-121=0). The monic equation is (x^2-(\)sum)x+product\(=0). Using sum (0) and product (-121) gives (x^2-121=0).\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) है। \(योग (0) और गुणनफल (-121) रखने पर (x^2-121=0) मिलता है\)।
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दो संख्याओं का योग (15) और गुणनफल (54) है। वे किस द्विघात समीकरण के मूल हो सकते हैं?
Two numbers have sum (15) and product (54). They can be roots of which quadratic equation?
#quadratic-equations
#forming-equation
#sum-product
#expert
A \(x^2-15x+54=0\)
B \(x^2+15x+54=0\)
C \(x^2-54x+15=0\)
D \(x^2+54x-15=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-15x+54=0\)
Step 1
Concept
If the sum of roots is (15) and product is (54), the equation is \(x^2-15x+54=0\). Remember the monic form formula.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-15x+54=0\). If the sum of roots is (15) and product is (54), the equation is \(x^2-15x+54=0\). Remember the monic form formula.
Step 3
Exam Tip
यदि मूलों का योग (15) और गुणनफल (54) है, तो समीकरण \(x^2-15x+54=0\) होगा। मोनिक रूप का सूत्र याद रखें।
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यदि किसी मोनिक द्विघात समीकरण के मूल (r) और (3r) हैं तथा उनका योग (16) है, तो उस समीकरण का स्थिर पद क्या होगा?
If the roots of a monic quadratic equation are (r) and (3r), and their sum is (16), what will be the constant term?
#quadratic-equations
#roots
#monic-equation
#expert
A (48)
B (16)
C (64)
D (32)
Explanation opens after your attempt
Step 1
Concept
From (r+3r=16), we get (r=4), so the roots are (4) and (12). The constant term is the product of roots (48).
Step 2
Why this answer is correct
The correct answer is A. (48). From (r+3r=16), we get (r=4), so the roots are (4) and (12). The constant term is the product of roots (48).
Step 3
Exam Tip
(r+3r=16) से (r=4) मिलता है, इसलिए मूल (4) और (12) हैं। स्थिर पद मूलों का गुणनफल (48) होगा।
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मूलों का योग (-11) और गुणनफल (30) वाला मोनिक द्विघात समीकरण कौन-सा है?
Which monic quadratic equation has sum of roots (-11) and product (30)?
#quadratic-equations
#sum-product
#forming-equation
#expert
A \(x^2+11x+30=0\)
B \(x^2-11x+30=0\)
C \(x^2+30x+11=0\)
D \(x^2-30x+11=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+11x+30=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Substituting sum (-11) gives (x^2+11x+30=0).\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2+11x+30=0). A monic equation is (x^2-(\)sum)x+product\(=0). Substituting sum (-11) gives (x^2+11x+30=0).\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(योग (-11) रखने पर (x^2+11x+30=0) मिलता है\)।
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किस द्विघात समीकरण के मूलों का योग (0) और गुणनफल (-81) है?
Which quadratic equation has sum of roots (0) and product (-81)?
#quadratic-equations
#sum-product
#forming-equation
#hard
A \(x^2-81=0\)
B \(x^2+81=0\)
C \(x^2+9x-81=0\)
D \(x^2-9x-81=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-81=0\)
Step 1
Concept
\(The monic equation is (x^2-(\)sum)x+product\(=0). Using sum (0) and product (-81) gives (x^2-81=0).\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-81=0). The monic equation is (x^2-(\)sum)x+product\(=0). Using sum (0) and product (-81) gives (x^2-81=0).\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) है। \(योग (0) और गुणनफल (-81) रखने पर (x^2-81=0) मिलता है\)।
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दो संख्याओं का योग (13) और गुणनफल (40) है। वे किस द्विघात समीकरण के मूल हो सकते हैं?
Two numbers have sum (13) and product (40). They can be roots of which quadratic equation?
#quadratic-equations
#forming-equation
#sum-product
#hard
A \(x^2-13x+40=0\)
B \(x^2+13x+40=0\)
C \(x^2-40x+13=0\)
D \(x^2+40x-13=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-13x+40=0\)
Step 1
Concept
If the sum of roots is (13) and product is (40), the equation is \(x^2-13x+40=0\). Remember the monic form formula.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-13x+40=0\). If the sum of roots is (13) and product is (40), the equation is \(x^2-13x+40=0\). Remember the monic form formula.
Step 3
Exam Tip
यदि मूलों का योग (13) और गुणनफल (40) है, तो समीकरण \(x^2-13x+40=0\) होगा। मोनिक रूप का सूत्र याद रखें।
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यदि किसी मोनिक द्विघात समीकरण के मूल (r) और (2r) हैं तथा उनका योग (9) है, तो उस समीकरण में स्थिर पद क्या होगा?
If the roots of a monic quadratic equation are (r) and (2r), and their sum is (9), what will be the constant term?
#quadratic-equations
#roots
#monic-equation
#hard
A (18)
B (9)
C (27)
D (36)
Explanation opens after your attempt
Step 1
Concept
From (r+2r=9), we get (r=3), so the roots are (3) and (6). The constant term will be the product of roots (18).
Step 2
Why this answer is correct
The correct answer is A. (18). From (r+2r=9), we get (r=3), so the roots are (3) and (6). The constant term will be the product of roots (18).
Step 3
Exam Tip
(r+2r=9) से (r=3) मिलता है, इसलिए मूल (3) और (6) हैं। स्थिर पद मूलों का गुणनफल (18) होगा।
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मूलों का योग (-9) और गुणनफल (20) वाला मोनिक द्विघात समीकरण कौन-सा है?
Which monic quadratic equation has sum of roots (-9) and product (20)?
#quadratic-equations
#sum-product
#forming-equation
#hard
A \(x^2+9x+20=0\)
B \(x^2-9x+20=0\)
C \(x^2+20x+9=0\)
D \(x^2-20x+9=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+9x+20=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Substituting sum (-9) gives (x^2+9x+20=0).\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2+9x+20=0). A monic equation is (x^2-(\)sum)x+product\(=0). Substituting sum (-9) gives (x^2+9x+20=0).\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(योग (-9) रखने पर (x^2+9x+20=0) मिलता है\)।
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किस द्विघात समीकरण के मूलों का योग (0) और गुणनफल (-49) है?
Which quadratic equation has sum of roots (0) and product (-49)?
#quadratic-equations
#sum-product
#forming-equation
#hard
A \(x^2-49=0\)
B \(x^2+49=0\)
C \(x^2+7x-49=0\)
D \(x^2-7x-49=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-49=0\)
Step 1
Concept
\(The monic equation is (x^2-(\)sum)x+product\(=0). Using sum (0) and product (-49) gives (x^2-49=0).\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-49=0). The monic equation is (x^2-(\)sum)x+product\(=0). Using sum (0) and product (-49) gives (x^2-49=0).\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) है। \(योग (0) और गुणनफल (-49) रखने पर (x^2-49=0) मिलता है\)।
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दो संख्याओं का योग (11) और गुणनफल (30) है। वे किस द्विघात समीकरण के मूल हो सकते हैं?
Two numbers have sum (11) and product (30). They can be roots of which quadratic equation?
#quadratic-equations
#forming-equation
#sum-product
#hard
A \(x^2-11x+30=0\)
B \(x^2+11x+30=0\)
C \(x^2-30x+11=0\)
D \(x^2+30x-11=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-11x+30=0\)
Step 1
Concept
If the sum of roots is (11) and product is (30), the equation is \(x^2-11x+30=0\). Remember the monic form formula.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-11x+30=0\). If the sum of roots is (11) and product is (30), the equation is \(x^2-11x+30=0\). Remember the monic form formula.
Step 3
Exam Tip
यदि मूलों का योग (11) और गुणनफल (30) है, तो समीकरण \(x^2-11x+30=0\) होगा। मोनिक रूप का सूत्र याद रखें।
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मूलों का योग (7) और गुणनफल (10) वाला मोनिक द्विघात समीकरण कौन-सा है?
Which monic quadratic equation has sum of roots (7) and product (10)?
#quadratic-equations
#sum-product
#forming-equation
#hard
A \(x^2-7x+10=0\)
B \(x^2+7x+10=0\)
C \(x^2-10x+7=0\)
D \(x^2+10x-7=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-7x+10=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-7x+10=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-7x+10=0). A monic equation is (x^2-(\)sum)x+product\(=0). Therefore (x^2-7x+10=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2-7x+10=0) सही है\)।
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किस द्विघात समीकरण में मूलों का योग (-8) और गुणनफल (15) है?
Which quadratic equation has sum of roots (-8) and product (15)?
#quadratic-equations
#sum-product
#forming-equation
#medium
A \(x^2+8x+15=0\)
B \(x^2-8x+15=0\)
C \(x^2+15x+8=0\)
D \(x^2-15x+8=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+8x+15=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Substituting sum (-8) gives (x^2+8x+15=0).\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2+8x+15=0). A monic equation is (x^2-(\)sum)x+product\(=0). Substituting sum (-8) gives (x^2+8x+15=0).\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(योग (-8) रखने पर (x^2+8x+15=0) मिलता है\)।
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यदि (a=-2), (b=9), (c=-7) हैं, तो संबंधित द्विघात समीकरण कौन-सा है?
If (a=-2), (b=9), (c=-7), which is the corresponding quadratic equation?
#quadratic-equations
#forming-equation
#coefficients
#medium
A \(2x^2+9x-7=0\)
B \(-2x^2+9x-7=0\)
C \(-2x^2-9x-7=0\)
D \(7x^2+9x-2=0\)
Explanation opens after your attempt
Correct Answer
B. \(-2x^2+9x-7=0\)
Step 1
Concept
Substituting in \(ax^2+bx+c=0\) gives \(-2x^2+9x-7=0\). Never ignore the sign of a coefficient.
Step 2
Why this answer is correct
The correct answer is B. \(-2x^2+9x-7=0\). Substituting in \(ax^2+bx+c=0\) gives \(-2x^2+9x-7=0\). Never ignore the sign of a coefficient.
Step 3
Exam Tip
मानक रूप \(ax^2+bx+c=0\) में मान रखने पर \(-2x^2+9x-7=0\) मिलता है। गुणांक का चिन्ह कभी न छोड़ें।
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किस द्विघात समीकरण में मूलों का योग (-6) और गुणनफल (8) है?
Which quadratic equation has sum of roots (-6) and product (8)?
#quadratic-equations
#sum-product
#forming-equation
#medium
A \(x^2+6x+8=0\)
B \(x^2-6x+8=0\)
C \(x^2+8x+6=0\)
D \(x^2-8x+6=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+6x+8=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Substituting sum (-6) gives (x^2+6x+8=0).\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2+6x+8=0). A monic equation is (x^2-(\)sum)x+product\(=0). Substituting sum (-6) gives (x^2+6x+8=0).\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(योग (-6) रखने पर (x^2+6x+8=0) मिलता है\)।
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यदि (a=5), (b=-11), (c=-6) हैं, तो संबंधित द्विघात समीकरण कौन-सा है?
If (a=5), (b=-11), (c=-6), which is the corresponding quadratic equation?
#quadratic-equations
#forming-equation
#coefficients
#medium
A \(5x^2+11x-6=0\)
B \(5x^2-11x-6=0\)
C \(6x^2-11x-5=0\)
D \(5x^2-6x-11=0\)
Explanation opens after your attempt
Correct Answer
B. \(5x^2-11x-6=0\)
Step 1
Concept
Substituting into \(ax^2+bx+c=0\) gives \(5x^2-11x-6=0\). Treat the negative sign as part of the coefficient.
Step 2
Why this answer is correct
The correct answer is B. \(5x^2-11x-6=0\). Substituting into \(ax^2+bx+c=0\) gives \(5x^2-11x-6=0\). Treat the negative sign as part of the coefficient.
Step 3
Exam Tip
मानक रूप \(ax^2+bx+c=0\) में मान रखने पर \(5x^2-11x-6=0\) मिलता है। ऋण चिन्ह को गुणांक का भाग मानें।
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किस द्विघात समीकरण में मूलों का योग (5) और गुणनफल (6) है?
Which quadratic equation has sum of roots (5) and product (6)?
#quadratic-equations
#sum-product
#forming-equation
#medium
A \(x^2-5x+6=0\)
B \(x^2+5x+6=0\)
C \(x^2-6x+5=0\)
D \(x^2+6x+5=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-5x+6=0\)
Step 1
Concept
\(A monic equation is (x^2-(\)sum)x+product\(=0). Hence (x^2-5x+6=0) is correct.\)
Step 2
Why this answer is correct
\(The correct answer is A. (x^2-5x+6=0). A monic equation is (x^2-(\)sum)x+product\(=0). Hence (x^2-5x+6=0) is correct.\)
Step 3
Exam Tip
\(मोनिक समीकरण (x^2-(\)योग)x+गुणनफल=0) होता है। \(इसलिए (x^2-5x+6=0) सही है\)।
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यदि (a=3,\ b=0,\ c=-12) हैं तो संबंधित द्विघात समीकरण कौन-सा है?
If (a=3,\ b=0,\ c=-12), which is the corresponding quadratic equation?
#quadratic-equations
#forming-equation
#coefficients
A \(3x^2-12=0\)
B \(3x^2+12=0\)
C \(3x^2-12x=0\)
D \(12x^2+3=0\)
Explanation opens after your attempt
Correct Answer
A. \(3x^2-12=0\)
Step 1
Concept
Substituting in \(ax^2+bx+c=0\) gives \(3x^2+0x-12=0\). Simplifying gives \(3x^2-12=0\).
Step 2
Why this answer is correct
The correct answer is A. \(3x^2-12=0\). Substituting in \(ax^2+bx+c=0\) gives \(3x^2+0x-12=0\). Simplifying gives \(3x^2-12=0\).
Step 3
Exam Tip
मानक रूप \(ax^2+bx+c=0\) में मान रखने पर \(3x^2+0x-12=0\) मिलता है। इसे सरल करने पर \(3x^2-12=0\) है।
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जिन मूलों में (0) और (-3) हों उनके लिए सही द्विघात समीकरण कौन-सा है?
Which quadratic equation has roots (0) and (-3)?
#quadratic-equations
#forming-equation
#roots
A \(x^2+3x=0\)
B \(x^2-3x=0\)
C \(x^2+3=0\)
D \(x^2-3=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+3x=0\)
Step 1
Concept
For roots (0) and (-3), the factors are (x) and (x+3). So the equation is (x(x+3)=0), that is \(x^2+3x=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+3x=0\). For roots (0) and (-3), the factors are (x) and (x+3). So the equation is (x(x+3)=0), that is \(x^2+3x=0\).
Step 3
Exam Tip
मूल (0) और (-3) होने पर गुणनखंड (x) और (x+3) होंगे। इसलिए समीकरण (x(x+3)=0) यानी \(x^2+3x=0\) है।
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समीकरण (3x+2=0) द्विघात क्यों नहीं है?
Why is (3x+2=0) not a quadratic equation?
#quadratic-equations
#non-example
#linear-equation
A क्योंकि इसमें \(x^2\) पद है / Because it has an \(x^2\) term
B क्योंकि इसमें स्थिर पद है / Because it has a constant term
C क्योंकि इसमें (x) नहीं है / Because it has no (x)
D क्योंकि सबसे बड़ी घात (1) है / Because the highest power is (1)
Explanation opens after your attempt
Correct Answer
D. क्योंकि सबसे बड़ी घात (1) है / Because the highest power is (1)
Step 1
Concept
In (3x+2=0), the highest power of (x) is (1). A quadratic equation must have highest power (2).
Step 2
Why this answer is correct
The correct answer is D. क्योंकि सबसे बड़ी घात (1) है / Because the highest power is (1). In (3x+2=0), the highest power of (x) is (1). A quadratic equation must have highest power (2).
Step 3
Exam Tip
(3x+2=0) में (x) की सबसे बड़ी घात (1) है। द्विघात के लिए सबसे बड़ी घात (2) होनी चाहिए।
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निम्न में से कौन-सा द्विघात समीकरण है?
Which of the following is a quadratic equation?
#quadratic-equations
#identify-equation
#class-10
#easy
A \(x^2+3x+2=0\)
B (2x+5=0)
C \(x^3-x=0\)
D \(\frac{1}{x}+2=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+3x+2=0\)
Step 1
Concept
In \(x^2+3x+2=0\), the highest power is (2). Check the power first while choosing an option.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+3x+2=0\). In \(x^2+3x+2=0\), the highest power is (2). Check the power first while choosing an option.
Step 3
Exam Tip
\(x^2+3x+2=0\) में सबसे बड़ी घात (2) है। विकल्प चुनते समय घात को सबसे पहले देखें।
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किस समीकरण में गुणनखंड विधि की जगह द्विघात सूत्र अधिक सुविधाजनक है?
For which equation is the quadratic formula more convenient than factorisation?
#quadratic
#method-selection
#quadratic-formula
A \(x^2+x-1=0\)
B \(x^2+5x+6=0\)
C \(x^2-9=0\)
D \(x^2-4x+4=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+x-1=0\)
Step 1
Concept
\(x^2+x-1=0\) has no simple integer factors, so the formula method is easier. In exams, the quadratic formula is safe in such cases.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+x-1=0\). \(x^2+x-1=0\) has no simple integer factors, so the formula method is easier. In exams, the quadratic formula is safe in such cases.
Step 3
Exam Tip
\(x^2+x-1=0\) के सरल पूर्णांक गुणनखंड नहीं मिलते, इसलिए सूत्र विधि आसान है। परीक्षा में ऐसे मामलों में द्विघात सूत्र सुरक्षित रहता है।
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किस विकल्प में दिया गया समीकरण द्विघात नहीं रहेगा?
In which option will the given equation not remain quadratic?
#quadratic-equations
#parameter
#not-quadratic
#hard
A ((t-2)x-2 +5x+1=0), (t=2)
B \(3x^2+tx+1=0\), (t=0)
C \(x^2+t=0\), (t=0)
D \(2x^2+1=0\), (t=2)
Explanation opens after your attempt
Correct Answer
A. ((t-2)x-2 +5x+1=0), (t=2)
Step 1
Concept
In the first option, putting (t=2) makes the coefficient of \(x^2\) equal to (0). Then the equation becomes linear.
Step 2
Why this answer is correct
The correct answer is A. ((t-2)x-2 +5x+1=0), (t=2). In the first option, putting (t=2) makes the coefficient of \(x^2\) equal to (0). Then the equation becomes linear.
Step 3
Exam Tip
पहले विकल्प में (t=2) रखने पर \(x^2\) का गुणांक (0) हो जाता है। तब समीकरण रैखिक बन जाता है।
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